[Math] Function for a free throw trajectory of ball

functionsquadratics

I've been trying to write functions for basketball's free throw trajectory.

These are main points:

  • the three throw line is 15 feet away from the foot of the basket
  • the basket is 10ft above the ground
  • the player releases the ball from approximate height of 8ft

I have to find two different functions for this information.

Let's denote the position of the basket as A, and the position of player as B, then:

$A(0;10)$ and $B(15;8)$.

First, the standard quadratic form:

$\ y = ax^{2} + bx + c$

Since 10 is the height required for the object to hit the basket –

$\ y = ax^{2} + bx + 10$

From my calculations, If c in function was greater or smaller than 10, Ball wouldn't hit the basket.

Then, considering that ball is shot from 8 feet and the distance from the starting point to the basket is 15 feet:

$\ 8 = a(15)^{2} + b(15) + 10$

Now to calculate the terms, I need to obtain the value of variable b, Thus, I get axis of symmetry, which is obviously 7.5.

Since:

$\ x[0] = 7.5$

$\frac{-b}{2a}=7.5$

$\ b=-15a$

Now substitution:

$\ 8 = 225a + (-15a)(15) + 10$

$\ 8 = 225a – 225a + 10$

$\ 8 = 10$

$\ 2$

I'm not sure afterwards, since i don't seem to properly obtain a, what could be the problem? Are the values that i have obtained correct?

So, what could be the proper function which would land the object in basket by it's graph?

Thanks!

Best Answer

There is one degree of freedom

Assume that the initial velocity of the throw is $p{\bf i} + q{\bf j}$, so $u_x=p$ and $u_y=q$. The acceleration due to gravity is $0{\bf i} - g{\bf j}$ , meaning that $a_x=0$ and $a_y=-g$. The throw point is 15ft from the basket, so $s_x=15$. The initial throw height is 8ft above the ground and the basket height is 10ft, so $s_y=2$.

This takes the throw point as the origin.

Considering only vertical motion: $s=ut+\frac{1}{2}at^2$ gives $15 = qt-\frac{1}{2}gt^2$.

Considering only horizontal motion: $s=ut+\frac{1}{2}at^2$ gives $2 = pt$.

Solving $2=pt$ gives $t=\frac{2}{p}$, assuming that $p \neq 0$, and so $15=q\left(\frac{2}{p}\right)-\frac{1}{2}g\left(\frac{2}{p}\right)^2$

$$15p^2=2pq-2g$$

$$q=\frac{15p^2+2g}{2p}$$

For any choice of $p$, you get a $q$ which gives the initial velocity.

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